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Since $e$ is a transcendence number, so it is certain that it is not zero of any polynomial with rational coefficients. However, I wonder can we find a power series with rational coefficient such that it is zero evaluated at $e$. If such series exists, can we explicitly find the coefficients?

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    $\begingroup$ Ethan Bolker's answer is great. I just wanted to comment on the last line, about series that are particularly nice. You are perhaps thinking about $\pi$ that is the zero of a particularly nice power series: the Taylor series of the sin function. Now one thing that sets $\pi$ apart from $e$ despite both of them being transcendental is that $\pi$ is a 'period' and $e$ conjecturally is not. I am inclined to believe b/c of this that no 'particularly nice' series for $e$ exists. The answers to my earlier MO question about this topic: mathoverflow.net/q/180035/41139 might be helpful here. $\endgroup$
    – Vincent
    Nov 25, 2021 at 16:05
  • $\begingroup$ This is, of course, all speculation $\endgroup$
    – Vincent
    Nov 25, 2021 at 16:08
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    $\begingroup$ Similar to Ethan Bolker's answer you can just use reciprocals so something like $1-\frac{x}{3}-\frac{x^2}{79}-\frac{x^3}{53748}-\ldots$ and even those terms have got you to below $10^{-8}$ and the next to below $10^{-19}$ $\endgroup$
    – Henry
    Nov 25, 2021 at 16:16

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Yes. You can build such a power series recursively.

Start with $a_0 = 1$ (or anywhere else you like).

Now find a rational multiple $a_1$ of $e$ such that $$ 0 < 1 + a_1e < 1/2. $$ Then find rational $a_2$ such that $$ 0 < 1 + a_1e + a_2e^2< 1/4. $$ Continue in the obvious way.

Clearly this procedure produces many such series. I don't know whether there's one that's particularly nice.

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